swift 在代码中将生成的 imageView 变成 UIButton

Question

我正在尝试向生成的图像添加点击手势，它工作正常，但一旦我点击图像，程序就会崩溃。我在这里做错了什么。任何帮助表示赞赏。

let tapGesture = UITapGestureRecognizer(target: self, action: "image Tapped")
            let imageView = UIImageView(image: UIImage(data: data));
            imageView.addGestureRecognizer(tapGesture)
            imageView.userInteractionEnabled = true

这是我得到的错误

2016-07-12 09:19:30.241 XXX[2170:27661] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: ‘-[MyViewController image Tapped]: unrecognized selector sent to instance 0x7fe811e30560'
 *** First throw call stack:
 (
0   CoreFoundation                      0x000000010e2a7d85 __exceptionPreprocess + 165
1   libobjc.A.dylib                     0x000000011144ddeb objc_exception_throw + 48
2   CoreFoundation                      0x000000010e2b0d3d -[NSObject(NSObject) doesNotRecognizeSelector:] + 205
3   CoreFoundation                      0x000000010e1f6cfa ___forwarding___ + 970
4   CoreFoundation                      0x000000010e1f68a8 _CF_forwarding_prep_0 + 120
5   UIKit                               0x0000000110391b28 _UIGestureRecognizerSendTargetActions + 153
6   UIKit                               0x000000011038e19a _UIGestureRecognizerSendActions + 162
7   UIKit                               0x000000011038c197 -[UIGestureRecognizer _updateGestureWithEvent:buttonEvent:] + 843
8   UIKit                               0x0000000110394655 ___UIGestureRecognizerUpdate_block_invoke898 + 79
9   UIKit                               0x00000001103944f3 _UIGestureRecognizerRemoveObjectsFromArrayAndApplyBlocks + 342
10  UIKit                               0x0000000110381e75 _UIGestureRecognizerUpdate + 2634
11  UIKit                               0x000000010ff0e48e -[UIWindow _sendGesturesForEvent:] + 1137
12  UIKit                               0x000000010ff0f6c4 -[UIWindow sendEvent:] + 849
13  UIKit                               0x000000010febadc6 -[UIApplication sendEvent:] + 263
14  UIKit                               0x000000010fe94553 _UIApplicationHandleEventQueue + 6660
15  CoreFoundation                      0x000000010e1cd301 __CFRUNLOOP_IS_CALLING_OUT_TO_A_SOURCE0_PERFORM_FUNCTION__ + 17
16  CoreFoundation                      0x000000010e1c322c __CFRunLoopDoSources0 + 556
17  CoreFoundation                      0x000000010e1c26e3 __CFRunLoopRun + 867
18  CoreFoundation                      0x000000010e1c20f8 CFRunLoopRunSpecific + 488
19  GraphicsServices                    0x0000000112d68ad2 GSEventRunModal + 161
20  UIKit                               0x000000010fe99f09 UIApplicationMain + 171
21  ApivitaClient                       0x000000010d6a1482 main + 114
22  libdyld.dylib                       0x0000000111f1192d start + 1
23  ???                                 0x0000000000000001 0x0 + 1

) libc++abi.dylib：以 NSException 类型的未捕获异常终止 (lldb)

Answer 1

替换

let tapGesture = UITapGestureRecognizer(target: self, action: "image Tapped")

与

let tapGesture = UITapGestureRecognizer(target:self, action: #selector(imageTapped))

并创建 func imageTapped() {}

Answer 2

UITapGestureRecognizer 初始化器中的动作部分表示将执行哪个方法。行动部分意味着谁负责该行动，谁是代表。

使用它来初始化手势识别器：

let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: Selector("imageTapped:"))

请注意方法名称后的冒号 : (imageTapped)。这意味着手势识别器将调用 imageTapped 方法并传递 sender 参数，如果您想确定触发了哪个手势识别器，这将很有用。如果您有多个可以点击的图像，这将非常有用，因此您不必为每个图像构建一个方法。

然后需要声明imageTapped方法：

func imageTapped(sender: UIGestureRecognizer) {
    print("Image tapped")
}

您也可以将函数声明为不带参数工作，只需声明不带 sender 参数的函数并删除操作选择器中的冒号即可。无论如何，我建议保留它，它可能在某个地方有用，如果你这样学习它会更好。

swift 在代码中将生成的 imageView 变成 UIButton

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